Problem: Simplify the following expression and state the condition under which the simplification is valid. $a = \dfrac{-n^2 + 10n - 21}{-9n^2 - 27n + 162}$
Answer: First factor out the greatest common factors in the numerator and in the denominator. $ a = \dfrac {-1(n^2 - 10n + 21)} {-9(n^2 + 3n - 18)} $ $ a = \dfrac{1}{9} \cdot \dfrac{n^2 - 10n + 21}{n^2 + 3n - 18} $ Next factor the numerator and denominator. $ a = \dfrac{1}{9} \cdot \dfrac{(n - 3)(n - 7)}{(n - 3)(n + 6)}$ Assuming $n \neq 3$ , we can cancel the $n - 3$ $ a = \dfrac{1}{9} \cdot \dfrac{n - 7}{n + 6}$ Therefore: $ a = \dfrac{ n - 7 }{ 9(n + 6)}$, $n \neq 3$